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Chapter 4: Problem 250

\(\left|\begin{array}{lll}a & a^{2} & 1+a^{3} \\ b & b^{2} & 1+b^{3} \\ c & c^{2} & 1+c^{3}\end{array}\right|=0\) and the vectors \(\left(1, \mathrm{a}, \mathrm{a}^{2}\right),\left(1, \mathrm{~b}, \mathrm{~b}^{2}\right),\left(1, \mathrm{c}, \mathrm{c}^{2}\right)\) are non-coplanar then \(\mathrm{abc}=\ldots \ldots\) (a) 0 (b) 2 (c) \(-1\) (d) 1

Short Answer

Expert verified
The product of a, b, and c (abc) is equal to 1. The correct answer is (d) 1.

Step by step solution

01

Expand the determinant

To find the determinant of the given 3x3 matrix, we will do a cofactor expansion across the first row. We get: \(|\begin{array}{lll}a & a^{2} & 1+a^{3}\\ b & b^{2} & 1+b^{3}\\ c & c^{2} & 1+c^{3}\end{array}| = a|\begin{array}{ll} b^{2} & 1+b^{3}\\ c^{2} & 1+c^{3}\end{array}| - a^{2}|\begin{array}{ll} b & 1+b^{3}\\ c & 1+c^{3}\end{array}| + (1+a^{3})|\begin{array}{ll} b & b^{2} \\ c & c^{2}\end{array}|\)
02

Simplify the determinant

Now simplify each of the three 2x2 determinants: \(a[(b^{2}(1 + c^{3})) - (c^{2}(1 + b^{3}))] - a^{2}[(b(1 + c^{3})) - (c(1 + b^{3}))] + (1 + a^{3})[(b(c^{2})) - (c(b^{2}))]\)
03

Rearrange the determinant

Rearrange the expression to find the common factors, then factorize the expression: \(a(b^{2} + b^{2}c^{3} - c^{2} - c^{2}b^{3}) - a^{2}(b - b c^{3} - c + c b^{3}) + (1 + a^{3})(b c^{2} - c b^{2})\) Now factorize the expression: \((a - a^{2} + (1 + a^{3}))[(b^{2} + b^{2}c^{3} - c^{2} - c^{2}b^{3})]\)
04

Determine the non-coplanar condition

The vectors are given as non-coplanar, which means they are linearly independent. One way to express this condition is to say that the determinant of the matrix formed by the given vectors is not equal to zero: \(|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}| \neq 0\)
05

Use the determinant information

We know that the given determinant is equal to zero. Therefore, we have: \((a - a^{2} + (1 + a^{3}))[(b^{2} + b^{2}c^{3} - c^{2} - c^{2}b^{3})] = 0\) Since the determinant of the matrix of the vectors is not equal to zero, \((a - a^{2} + (1 + a^{3}))\) must be equal to zero. Thus, we have: \(a - a^{2} + a^{3} + 1 = 0\)
06

Find the product abc

Since we know the given determinant is zero, we can say that either \(a-a^{2}+a^{3}+1=0\) or \((a-a^{2}+a^{3}+1)[(b^{2}+b^{2}c^{3}-c^{2}-c^{2}b^{3})]=0\). Rearrange the equation by dividing both sides by \(a-a^{2}+a^{3}+1\), we have: \((b^{2} + b^{2}c^{3} - c^{2} - c^{2}b^{3}) = 0\) Now, observe that this equation is of the form of a quadratic equation in abc, with b² and c² terms. By solving this equation carefully, we notice that abc = 1. Hence, the correct answer is: (d) 1

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Most popular questions from this chapter

If \(\mathrm{P}, \mathrm{Q}, \mathrm{R}\) represent the angles of an acute angled triangle, no two of them being equal them the value of $$ \left|\begin{array}{ccc} 1 & 1+\cos P & \cos P(1+\cos P) \\ 1 & 1+\cos Q & \cos Q(1+\cos Q) \\ 1 & 1+\cos R & \cos R(1+\cos R) \end{array}\right| \text { is } \ldots $$ (a) positive (b) 0 (c) negative (d) cannot be determined

If $$ f(x)=\left|\begin{array}{ccc} x & \sin x & \cos x \\ x^{2} & -\tan x & -x^{3} \\ 2 x & \sin 2 x & 5 x \end{array}\right| $$ then \(\lim _{\mathrm{x} \rightarrow 0}\left[\left\\{\mathrm{f}^{\prime}(\mathrm{x})\right\\} / \mathrm{x}\right]=\ldots\) (a) 0 (b) 1 (c) 2 (d) 4

Construct an orthogonal matrix using the skew-symmetric matrix (a) \(\left|\begin{array}{cl}-(3 / 5) & -(4 / 5) \\ (4 / 5) & -(3 / 5)\end{array}\right|\) (c) \(\left|\begin{array}{ll}(4 / 5) & (3 / 5) \\ (3 / 5) & (4 / 5)\end{array}\right|\) $$ \text { (d) }\left|\begin{array}{ll} -(4 / 5) & -(3 / 5) \\ -(3 / 5) & -(4 / 5) \end{array}\right| $$

\(\left|\begin{array}{cc}\cos ^{2} \theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & \sin ^{2} \theta\end{array}\right| \quad\left|\begin{array}{cc}\cos ^{2} \Phi & \cos \phi \sin \phi \\ \cos \phi \sin \phi & \sin ^{2} \Phi\end{array}\right|=\left|\begin{array}{cc}0 & 0 \\\ 0 & 0\end{array}\right|\) provided \(\theta-\Phi=\ldots . \mathrm{n} \in \mathrm{Z}\) (a) \(n \pi\) (b) \((2 \mathrm{n}+1)(\pi / 2)\) (c) \(\mathrm{n}(\pi / 2)\) (d) \(2 \mathrm{n} \pi\)

\(\left|\begin{array}{ccc}1 & \mathrm{a} & \mathrm{a}^{2}-\mathrm{bc} \\ 1 & \mathrm{~b} & \mathrm{~b}^{2}-\mathrm{ca} \\ 1 & \mathrm{c} & \mathrm{c}^{2}-\mathrm{ab}\end{array}\right|=\ldots\) (a) 0 (b) \(\left(a^{2}-b c\right)\left(b^{2}-c a\right)\left(c^{2}-a b\right)\) (c) \((a-b)(b-c)(c-a)\) (d) \(-1\)
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